An object weighing 0.09lbs is connected to a light durablestring. The object is

Question

An object weighing 0.09lbs is connected to a light durablestring. The object is spun with circular trajectory horizontal tothe floor. The string dips a constant 12degrees from the horizontaland has a lenght of 60cm from its pivot point. Determine thefollowing. a) The linear speed V of the object b) The angular speed of the object (in rpm) c) The tension in the string An object weighing 0.09lbs is connected to a light durablestring. The object is spun with circular trajectory horizontal tothe floor. The string dips a constant 12degrees from the horizontaland has a lenght of 60cm from its pivot point. Determine thefollowing. a) The linear speed V of the object b) The angular speed of the object (in rpm) c) The tension in the string

Explanation / Answer

The mass of the object is m = 0.09 lbs = 0.09 * 0.4536 kg =0.041 kg The angle of dip of the string from the horizontal is =12o The length of the string from its pivot point is l = 60 cm =60 * 10-2 m a)The centripetal force acting on the string is equal to theweight of the string therefore we get (m * v2/l) = W * sin = mg *sin or v = (g * l* sin)1/2 -----------(1) g = 9.8 m/s^2 b)The angular speed of the object is v = l * w or w = (v/l) radians/second = (v/l) * (60/2) rpm the value of v is obtained from equation (1) c)The tension in the string is equal to the centripetal forceacting on the string therefore we get T = (m * v2/l) g = 9.8 m/s^2 b)The angular speed of the object is v = l * w or w = (v/l) radians/second = (v/l) * (60/2) rpm the value of v is obtained from equation (1) c)The tension in the string is equal to the centripetal forceacting on the string therefore we get T = (m * v2/l)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.